Number of correct bond angle order out of the following are ( X = central atom) (i) NH 4 + PH 3 AsH 3

Question

Number of correct bond angle order out of the following are ( $X =$ central atom) (i) $NH_{4}^{+} < NH_{3} < NH_{2}^{-} (H - \hat{N} - H)$ (ii) $NO_{2}^{+} > NO_{2}^{-} (O - \hat{N} - O)$ (iii) $NH_{3} > PH_{3} > AsH_{3} (H - \hat{X} - H)$ (iv) $PCl_{4}^{+} < PBr_{4}^{+} < PI_{4}^{+} (X^{'} - \hat{P} - X^{'})$ Where $X^{'} =$ halogen (v) $CH_{4} > NH_{3} > H_{2} O (H - \hat{X} - H)$ (vi) $NH_{3} > H_{2} O (H - \hat{X} - H)$ (vii) $XeO_{4} = SiCl_{4} = SO_{4}^{- 2}$ (same bond angle)

TestHub · Accepted Answer

The total number of correct bond angle orders is 5. Statements (ii), (iii), (v), (vi), and (vii) are correct.

(ii), (v), (vi): Increasing lone pairs increases repulsion, decreasing the bond angle ( $N O_{2}^{+} = 18 0^{\circ}$ , $C H_{4} = 109. 5^{\circ} > N H_{3} = 10 7^{\circ} > H_{2} O = 104. 5^{\circ}$ ).

(iii): Decreasing electronegativity of the central atom down the group reduces bond-pair repulsion, lowering the angle ( $N H_{3} > P H_{3} > A s H_{3}$ ).

(vii): $X e O_{4}$ , $S i C l_{4}$ , and $S O_{4}^{2 -}$ are all perfectly symmetrical tetrahedral structures ( $s p^{3}$ , 0 lone pairs) with equal angles ( $109. 5^{\circ}$ ).

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